summer 2020 first run

Author: teepeemm

apexNotes.txt

@@ -1,3 +1,6 @@
+"chapter" "section"
+5.2 p207: should y be v in figures?
+
 Ally complains that "This HTML file contains an invalid heading structure"
 (image?) directories in Windows use a separator \ which becomes %5C in html, and breaks
 
@@ -10,6 +13,8 @@ from pretext/
 xsltproc -xinclude ../../mathbook/xsl/mathbook-html.xsl index.html ../xmlonly/index.xml
 
 
+add hierarchy of function growths to LH section?
+
 
 Instead of “Note: these are the same functions as in Exercises ##—##”, should we just combine them into a single problem with multiple parts?
 
@@ -84,10 +89,20 @@ My to do:
   vspace to addvspace, [small|med|big]skip to [small|med|big]break ?
 
 Typos:
+
+1.4p32 Th1.4.1: This needs to include "except possibly at $c$".
+3.3p143 KI3.3.1: Since 1. allows for f' to be undefined, we should just have f be a continuous function, not necessarily a differentiable one.
+5.4p236 T5.4.1: In order to be used by T5.4.4, you also need to conclude that $F$ is continuous at the endpoints.
+6.7p340#42: 1^+ should be 1^-
 13.7p837#10: Cylindrical should  be spherical.
+13.7p837#15 solution: $z=1$ should be $z=0$.
 13.7p838#17-22: The differentials should be in the order $d\rho d\phi d\theta$.
 14.2p855 line -5: *Will* all the loops of different directions in the field,
+14.3p861 Example 14.3.1 solution line 5: We should have $\vec r'_1(t)=\bracket{1,1}$, not $\bracket{1,2}$.
+14.3p869#17-20: The solutions use (a)-(d); the problem directions use 1.-4.
 14.4p879#19: The given curve is not closed. Using $\vec r(t)=\bracket{3t^2-2t-t^3,2(t-1)^2}$ gives the book's solution.
+14.5p890#9: For $x=1$, the $x$-component to $\vec r$ should be 1.
 14.5p890#13: The vertical surface is x^2+y^2/9=1 (3 times).
 14.5p890#17: $2\le v\le3$ should be $2\le y\le3$.
+14.5p890#23 solution: $\int_{-1}^1\int_0^{1-v^2}\sqrt{27}\ du\ dv=4\sqrt3\approx6.928$
 14.6p899#5: The mass function should be $\delta=z+10$ in order to stay positive and give the book's answer.

changes.md

@@ -1,6 +1,17 @@
 List Of Changes
 =====================
 
+Changes for the 2020-06 version:
+--------------------
+
+* A careful reading of Definition 3.3.1 shows that intervals of increasing and decreasing are usually closed within the domain.  We often had open intervals instead.
+
+* Definition 10.3.1 (Tangent line to a parametric curve) has been rearranged to no longer have an unnecessary "provided x'≠0".
+
+* Anthony Bevelacqua and John Collings contributed several exercises.
+
+* All known errata (20+) have been fixed.
+
 Changes for the 2019-06 version:
 --------------------
 

errata/Errata.tex

@@ -4,7 +4,8 @@
 \usepackage{booktabs}
 \usepackage{url}
 
-\title{Errata and Addenda to Apex LT}
+\title{Errata %and Addenda
+to Apex LT}
 \date{\today}
 
 \newcommand{\ds}{\displaystyle}
@@ -40,25 +41,56 @@
 \maketitle
 
 \noindent
-The following errors exist in the in June 2019 printed version  of Apex LT Calculus II:%\\[-2\baselineskip]
+The following errors exist in the in June 2019 printed version of Apex LT Calculus I:%\\[-2\baselineskip]
 \begin{enumerate}
+\item \S1.4 p41: Theorem 1.4.1 needs to say ``except possibly at $c$''.
+\item \S1.6\#14,16: To be continuous at a point, the function needs to be defined in a neighborhood of the point.  This means that the functions are not continuous at the indicated points.
+\item \S3.3: A careful reading of Definition 3.3.1 shows that intervals of increasing and decreasing are usually closed.  We often had open intervals instead.
+\item \S5.4 p262: In order for Theorem 5.4.1 (FTC1) to help with the MVT later on, this needs to also include that $F$ is continuous at the endpoints.
+\item In the integration formulas at the back of the book, \#13 is missing its $dx$.
+\label{2019-06-00Iplus}
+\end{enumerate}
+
+\noindent
+The following errors exist in the in June 2019 printed version of Apex LT Calculus II:%\\[-2\baselineskip]
+\begin{enumerate}
+\item \S7.4 p364 Line -1: as $x\to\infty$, both $\sinh x$ and $\cosh x$ approach $e^x/2$.
+\item \S7.5 p382\#40: $\ds\lim_{x\to1^+}$ should be $\ds\lim_{x\to1^-}$.
+\item \S8.6 p438 Figure 8.6.3: The graph is $f(x)=1/x$, not $1/x^2$.
 \item The chapter headings in \S10.2--10.5 show Chapter 9. % todo
-\label{2019-06-00IIplus}
+\item \S10.3 p608\#40: In order to be integrable, we should have $y=4e^{t/2}$.  The length is then $e^3+11-e^{-8}$.
+\item In the integration formulas at the back of the book, \#13 is missing its $dx$.
+\label{2019-06-00II}
 \end{enumerate}
 
 \noindent
-The following errors exist in the in June 2019 printed version  of Apex LT Calculus III:%\\[-2\baselineskip]
+The following errors exist in the in June 2019 printed version of Apex LT Calculus III:%\\[-2\baselineskip]
 \begin{enumerate}
+\item \S11.2 p659 Definition 11.2.3 refers to $c\vec v$ as a scalar product, whereas most authors use ``scalar product'' as a synonym for the dot product.
+\item \S11.2 p660 Line 17: This is the definition of $\vec u-\vec v$; it does not follow from anything.
+\item \S12.1 p720\#29: In the answer, the $z$-component should be $ht$.
+\item \S13.5 p817\#3: We need to state how the functions are defined.
+\item \S13.7 p839\#24: The equation should be $\sin(xy)+\cos(yz)=1$.
+\item \S13.8 Definition 13.8.1 part 1: We need $D\subset S$.
+\item \S14.5: We need to refer to the surface area of $z=f(x,y)$, not $f(x,y)$.
+\item \S14.6 Example 4: The definition of $D$ should clarify that we want $0\le z\le-y$.
 \item \S14.7 p944\#12: Cylindrical should be Spherical.
+\item \S14.7 p944\#15 solution: $z=1$ should be $z=0$.
 \item \S14.7 p945\#19--24: The differentials should be in the order $d\rho d\phi d\theta$.
+\item \S15.3 p970 Line 8: We should have $\vec r'_1(t)=\bracket{1,1}$, not $\bracket{1,2}$.
 \item \S15.4 p991\#31: The curve should be given by $\vec r(t)=\bracket{3t^2-2t-t^3,2(t-1)^2}$.
+\item \S15.5 p1002\#9 solution: For $x=1$, the $x$-component to $\vec r$ should be 1.
 \item \S15.5 p1002\#13: The vertical surface is given by $x^2+y^2/9=1$.
 \item \S15.5 p1003\#17: $2\le v\le3$ should be $2\le y\le3$.
+\item \S15.5 p1003\#27 solution: The integral should be $\int_{-1}^1\int_0^{1-v^2}\sqrt{27}\ du\ dv$. The exact value of the integral is $4\sqrt3\approx6.928$.
 \item \S15.6 p1012\#5: The mass function should be $\delta(x,y,z)=z+10$ in order to give the book's answer and maintain a positive mass density.
-\label{2019-06-00III}
+\item \S15.7 p1024 solution: The flux through the plane is 22.
+\item In the integration formulas at the back of the book, \#13 is missing its $dx$.
+\label{2019-06-00IIIplus}
 \end{enumerate}
 
 \begin{table}[h]
+\caption{Errata Tally (``+'' indicates systemic errata)}
 \begin{tabular}{lccc}\toprule
 Version & Calculus I & Calculus II & Calculus III \\\midrule
 \errorrow{2019-06-00} \\
@@ -70,10 +102,9 @@
 \errorrow{2016-08-00} \\
 \bottomrule
 \end{tabular}
-\caption{Errata Tally (``+'' indicates systemic errata)}
 \end{table}
 
-\newpage
+%\newpage
 
 \noindent
 The following errors exist in the July 13, 2018 printed version of Apex LT Calculus in the Important Formulas at the end of the book:
@@ -202,7 +233,7 @@
 \label{2018-07-13IIplus}
 \end{enumerate}\bigskip
 
-In addition to 1--\ref{2018-07-13}, the following error exists in the July 13, 2018 printed version of Apex LT Calculus III:
+In addition to 1--\ref{2018-07-13}, the following errors exist in the July 13, 2018 printed version of Apex LT Calculus III:
 \begin{enumerate}\setcounter{enumi}{\reforzero{2018-07-13}}
 \item It turns out that the units newton and joule are not capitalized.
 \item \S11.1\#29,30: The subsequent image has clipped the problem statement.\\
@@ -226,7 +257,7 @@ \section*{Digital Math Resources}
 
 %\end{document}
 
-\newpage
+%\newpage
 
 In the July 27, 2017 and November 13, 2017 printed versions of Apex LT Calculus, there are numerous instances of ``$\lim A+B$''.  The convention seems to be that this should be ``$\lim(A+B)$''.  We believe that all such instances have been corrected for subsequent versions of the text.\bigskip
 
@@ -263,7 +294,7 @@ \section*{Digital Math Resources}
 \label{2017-07-27Iplus}
 \end{enumerate}
 
-\clearpage
+%\clearpage
 
 The following errors exist in the July 27, 2017 printed versions of Apex LT Calculus II:
 \begin{enumerate}
@@ -308,7 +339,7 @@ \section*{Digital Math Resources}
 
 %\end{document}
 
-\clearpage
+%\clearpage
 
 In addition to the previous, the following errors exist in the July 27, 2017 printed version of Apex LT Calculus III:
 \begin{enumerate}
@@ -325,7 +356,7 @@ \section*{Digital Math Resources}
 \label{2017-07-27III}
 \end{enumerate}
 
-\clearpage
+%\clearpage
 
 In addition to the previous, the following error exists in the Summer 2017 printed version of Apex LT Calculus I:
 \begin{enumerate}
@@ -341,7 +372,7 @@ \section*{Digital Math Resources}
 \label{2017-05-00II}
 \end{enumerate}
 
-\clearpage
+%\clearpage
 
 In addition to the previous, the following errors exist in the Summer 2017 printed version of Apex LT Calculus III:
 \begin{enumerate}
@@ -391,7 +422,7 @@ \section*{Digital Math Resources}
 \label{2017-05-00IIIplus}
 \end{enumerate}
 
-\clearpage
+%\clearpage
 
 \newcommand{\springerrors}{%
 In addition to the previous, the following errors exist in the Spring 2017 printed version of Apex LT:
@@ -416,7 +447,7 @@ \section*{Digital Math Resources}
 \label{2017-01-00Iplus}
 \end{enumerate}
 
-\clearpage
+%\clearpage
 
 \springerrors
 
@@ -445,7 +476,7 @@ \section*{Digital Math Resources}
 \label{2017-01-00IIplus}
 \end{enumerate}
 
-\clearpage
+%\clearpage
 
 In addition to most of the previous, the following errors exist in the Fall 2016 printed version of Apex LT:
 \begin{enumerate}

exercises/01_05_exset_01.tex

@@ -49,7 +49,7 @@
 \node [above] at (myplot.above origin) {\scriptsize $y$};
 \end{tikzpicture}}{No; $f(1)$ does not exist.}
 
-\exercise{\noindent $a = 0$\\
+\exercise{\noindent $a = 1$\\
 \begin{tikzpicture}
 \begin{axis}[width=\marginparwidth,tick label style={font=\scriptsize},
 axis y line=middle,axis x line=middle,name=myplot,
@@ -64,7 +64,7 @@
 \end{axis}
 \node [right] at (myplot.right of origin) {\scriptsize $x$};
 \node [above] at (myplot.above origin) {\scriptsize $y$};
-\end{tikzpicture}}{Yes}
+\end{tikzpicture}}{No}
 
 \exercise{\noindent $a = 1$\\
 \begin{tikzpicture}
@@ -81,13 +81,13 @@
 \node [above] at (myplot.above origin) {\scriptsize $y$};
 \end{tikzpicture}}{Yes}
 
-\exercise{\noindent $a = 4$\\
+\exercise{\noindent $a = 2$\\
 \begin{tikzpicture}
 \begin{axis}[width=\marginparwidth,tick label style={font=\scriptsize},
 axis y line=middle,axis x line=middle,name=myplot,xtick={-4,...,-1,1,2,...,4},
 ymin=-4.5,ymax=4.5,xmin=-4.5,xmax=4.5]
-\addplot [thick,draw={\colorone},domain=-4:0] {4*cos(deg(x)*3.14159/4};
-\addplot [thick,draw={\colorone},domain=0:4]  {-4*cos(deg(x)*3.14159/4};
+\addplot [thick,draw={\colorone},domain=-4:0] {4*cos(45*x)};
+\addplot [thick,draw={\colorone},domain=0:4]  {-4*cos(45*x)};
 \fill[black,draw=black] (axis cs:0,0) circle (1.5pt);
 \fill[white,draw=black] (axis cs:0,4) circle (1.5pt);
 \fill[black,draw=black] (axis cs:-4,-4) circle (1.5pt);

exercises/03_03_exercises.tex

@@ -24,8 +24,8 @@
 \end{axis}
 \node [right] at (myplot.right of origin) {\scriptsize $x$};
 \node [above] at (myplot.above origin) {\scriptsize $y$};
-\end{tikzpicture}}{decreasing on $(-3,-1)$; $(1,3)$,\\
-increasing on $(-\infty,-3)$; $(-1,1)$; $(3,\infty)$;\\
+\end{tikzpicture}}{decreasing on $[-3,-1]$; $[1,3]$,\\
+increasing on $(-\infty,-3]$; $[-1,1]$; $[3,\infty)$;\\
 local maxima when $x=-3,1$,\\
 local minima when $x=-1,3$.}
 
@@ -41,8 +41,8 @@
 \end{axis}
 \node [right] at (myplot.right of origin) {\scriptsize $x$};
 \node [above] at (myplot.above origin) {\scriptsize $y$};
-\end{tikzpicture}}{decreasing on $(0,\frac\pi6)$; $(\frac\pi2,\frac{5\pi6})$; $(\frac{3\pi}2,2\pi)$,\\
-increasing on $(\frac\pi6,\frac\pi2)$; $(\frac{5\pi}6,\frac{3\pi}2)$;\\
+\end{tikzpicture}}{decreasing on $[0,\frac\pi6]$; $[\frac\pi2,\frac{5\pi6}]$; $[\frac{3\pi}2,2\pi]$,\\
+increasing on $[\frac\pi6,\frac\pi2]$; $[\frac{5\pi}6,\frac{3\pi}2]$;\\
 local maxima when $x=\frac\pi2,\frac{3\pi}2$,\\
 local minima when $x=\frac\pi6,\frac{5\pi}6$.}
 
@@ -56,8 +56,8 @@
 \end{axis}
 \node [right] at (myplot.right of origin) {\scriptsize $x$};
 \node [above] at (myplot.above origin) {\scriptsize $y$};
-\end{tikzpicture}}{decreasing on $(-\infty,-2)$; $(2,\infty)$,\\
-increasing on $(-2,2)$;\\
+\end{tikzpicture}}{decreasing on $(-\infty,-2]$; $[2,\infty)$,\\
+increasing on $[-2,2]$;\\
 local maxima when $x=2$,\\
 local minima when $x=-2$.}
 
@@ -71,8 +71,8 @@
 \end{axis}
 \node [right] at (myplot.right of origin) {\scriptsize $x$};
 \node [above] at (myplot.above origin) {\scriptsize $y$};
-\end{tikzpicture}}{decreasing on $(-1,1)$,\\
-increasing on $(-\infty,-1)$; $(1,\infty)$;\\
+\end{tikzpicture}}{decreasing on $[-1,1]$,\\
+increasing on $(-\infty,-1]$; $[1,\infty)$;\\
 local maxima when $x=-1$,\\
 local minima when $x=1$.}
 

exercises/03_03_exset_01.tex

@@ -1,39 +1,23 @@
 \exerciseset{In Exercises}{, a function $f(x)$ is given.
-	\begin{enumerate}
-	\item	[(a)] Compute $\fp(x)$.
-	\item	[(b)] Graph $f$ and $\fp$ on the same axes (using technology is permitted) and verify \autoref{thm:incr_decr}.
-	\end{enumerate}
-}{
-
-\exercise{$f(x) = 2x+3$
-}{Graph and verify.
-}
+	\begin{enumerate}[label=(\alph*)]
+	\item	Compute $\fp(x)$.
+	\item	Graph $f$ and $\fp$ on the same axes (using technology is permitted) and verify \autoref{thm:incr_decr}.
+	\end{enumerate}}{
 
-\exercise{$f(x) = x^2-3x+5$
-}{Graph and verify.
-}
+\exercise{$f(x) = 2x+3$}{Graph and verify.}
 
-\exercise{$f(x) = \cos x$
-}{Graph and verify.
-}
+\exercise{$f(x) = x^2-3x+5$}{Graph and verify.}
 
-\exercise{$f(x) = \tan x$
-}{Graph and verify.
-}
+\exercise{$f(x) = \cos x$}{Graph and verify.}
 
-\exercise{$f(x) = x^3-5x^2+7x-1$
-}{Graph and verify.
-}
+\exercise{$f(x) = \tan x$}{Graph and verify.}
 
-\exercise{$f(x) = 2x^3-x^2+x-1$
-}{Graph and verify.
-}
+\exercise{$f(x) = x^3-5x^2+7x-1$}{Graph and verify.}
 
-\exercise{$f(x) =x^4-5x^2+4$
-}{Graph and verify.
-}
+\exercise{$f(x) = 2x^3-x^2+x-1$}{Graph and verify.}
+
+\exercise{$f(x) =x^4-5x^2+4$}{Graph and verify.}
+
+\exercise{$\ds f(x) =\frac{1}{x^2+1}$}{Graph and verify.}
 
-\exercise{$\ds f(x) =\frac{1}{x^2+1}$
-}{Graph and verify.
 }
-}